Question

There are counters available in 3 different colours. The counters are all alike except for the colour. The total number of arrangement consisting of 5 counters, assuming sufficient number of counters of each colour if each arrangement consists of counters atleast one of each colour is :

• A. ${}^5{C}_3\cdot 3!\cdot 3^2$
• B. $3^5-\left({}^3{C}_2\cdot 2^5+{}^3{C}_1\right)$
• C. $3\left(3^4-2^5\right)$
• D. ${}^3{C}_1\cdot \frac{5!}{3!}+{}^3{C}_2+\frac{5!}{2!2!}$

Answer 1

Answer: (D)

A A A ............, B B B ............., C C C .............
$XXXXX;3^5-[$ No. of ways when exactly one colour counter is not there + when exactly two colour counters are not there]
$=3^5-\left[{}^3{C}_1\left(2^5-2\right)+{}^3{C}_2\right]$ ......(1)
altemately
A B C X X
we can choose both alike counters for two crosses or we can choose both different.
Accordingly
${}^3{C}_1\cdot \frac{5!}{3!}+{}^3{C}_2\cdot \frac{5!}{2!2!}$ ............(2)
(1) and (2) gives $150\Rightarrow D$